 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  I've a broken stick (Posted on 2003-12-07) I've a straight stick which has been broken into three random-length pieces.
What is the probability that the pieces can be put together to form a triangle?
If you can answer this at this point, please do.
If not, perhaps this will help: here are several methods to break the stick into the three random length pieces:
1. I select, independently, and at random, two points from the points that range uniformly along the stick, then break the stick at these two points.
2. I select one point, independently, and at random (again uniformly), and break the stick at this point. I then randomly (with even chances) select one of the two sticks and randomly select a point (again uniformly) along that stick, and break it at that point.
3. I select one point, independently, and at random (again uniformly), and break the stick at this point. I then select the larger stick, and randomly select a point (again uniformly) along that stick, and break it at that point.
If this clarifies the problem, please show how this affects your work.

 No Solution Yet Submitted by SilverKnight Rating: 3.6000 (5 votes) Comments: ( Back to comment list | You must be logged in to post comments.) re: I think this is a solution | Comment 19 of 26 | (In reply to I think this is a solution by puzzlesrfun)

"The sides of these smaller triangles are given by sqrt(2)*(1/4)*L. The 1/4 follows from the fact that two of the components are less than or equal to (1/4)*L, and therefore must sum to less than (1/2)*L."

But, actually, two of the components can add up to less than (1/2)*L even if not both of them are less than (1/4)*L. For instance, one is (1/8)*L and the other is (5/16)*L: the total is still less than half L. The better way to look at it is that no one component can be more than (1/2) L. The side of each small triangle is sqrt(2)*L/2.

As the larger triangle has area (sqrt(3)/2)*L^2 and the small triangles have area (sqrt(3)/2)*(L/2)^2, we get

(sqrt(3)/2)*L^2 - 3*(sqrt(3)/2)*(L/2)^2
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(sqrt(3)/2)*L^2

or 1 - 3/4 = 1/4.

 Posted by Charlie on 2003-12-10 14:03:03 Please log in:

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